Simplify and expand the following expression: $ \dfrac{2}{5y + 15}- \dfrac{1}{y - 3}+ \dfrac{2y}{y^2 - 9} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{2}{5y + 15} = \dfrac{2}{5(y + 3)}$ We can factor the quadratic in the third term: $ \dfrac{2y}{y^2 - 9} = \dfrac{2y}{(y + 3)(y - 3)}$ Now we have: $ \dfrac{2}{5(y + 3)}- \dfrac{1}{y - 3}+ \dfrac{2y}{(y + 3)(y - 3)} $ The least common multiple of the denominators is: $ 5(y + 3)(y - 3)$ In order to get the first term over $5(y + 3)(y - 3)$ , multiply by $\dfrac{y - 3}{y - 3}$ $ \dfrac{2}{5(y + 3)} \times \dfrac{y - 3}{y - 3} = \dfrac{2(y - 3)}{5(y + 3)(y - 3)} $ In order to get the second term over $5(y + 3)(y - 3)$ , multiply by $\dfrac{5(y + 3)}{5(y + 3)}$ $ \dfrac{1}{y - 3} \times \dfrac{5(y + 3)}{5(y + 3)} = \dfrac{5(y + 3)}{5(y + 3)(y - 3)} $ In order to get the third term over $5(y + 3)(y - 3)$ , multiply by $\dfrac{5}{5}$ $ \dfrac{2y}{(y + 3)(y - 3)} \times \dfrac{5}{5} = \dfrac{10y}{5(y + 3)(y - 3)} $ Now we have: $ \dfrac{2(y - 3)}{5(y + 3)(y - 3)} - \dfrac{5(y + 3)}{5(y + 3)(y - 3)} + \dfrac{10y}{5(y + 3)(y - 3)} $ $ = \dfrac{ 2(y - 3) - 5(y + 3) + 10y} {5(y + 3)(y - 3)} $ Expand: $ = \dfrac{2y - 6 - 5y - 15 + 10y}{5y^2 - 45} $ $ = \dfrac{7y - 21}{5y^2 - 45}$